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Aadarsh

  • 2 years ago

Important Question. Please explain with steps.

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  1. Aadarsh
    • 2 years ago
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  2. Aadarsh
    • 2 years ago
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    @Mani_Jha , @saifoo.khan , @heena , @supercrazy92 please help.

  3. saifoo.khan
    • 2 years ago
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    No chem! :(

  4. Aadarsh
    • 2 years ago
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    Try at least.

  5. saifoo.khan
    • 2 years ago
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    @Mani_Jha is a boss at chem.

  6. Aadarsh
    • 2 years ago
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    Yeah, that's why i called Mani Bhai also. He is my teacher. I hope he will do it this time too.:)))))

  7. Mani_Jha
    • 2 years ago
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    Let the concentration of NH3 be x. Now, the resultant solution was basic(as it had to be neutralized by NaOH). During the reaction of neutralization: M1V1=M2V2 x20=30*0.1 Find z. Let the initial volume of NH3 be V. Then when NH3 reacted initially with H2S04, 100 ml of H2S04 neutralized (v-20) ml of NH3(Because 20 ml was un-neutralized and made the solution basic) Again use the law of equivalents in this case(be careful: the n-factor of H2S04 is 2) Did you understand? @Saifoo.khan, you just guessed that, didnt u?

  8. Aadarsh
    • 2 years ago
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    Mani Bhai, can u show the second equation what we need to take?

  9. Mani_Jha
    • 2 years ago
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    (v-20)x=100*0.1*2. ok?

  10. Aadarsh
    • 2 years ago
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    Okay, mani bhaiiiiiii

  11. Vincent-Lyon.Fr
    • 2 years ago
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    Wordings of these problems are terrible; teachers should bear in mind that students are not reading their minds and that wordings should be unambiguous. In first problem, I find 17 millimole NH3 or 0.38 litre in STP. That holds if "equivalence" in the wording is meant for 'first equivalence', because eventually, NH4+ formed will also be titrated. In second problem there should be 1.5 millimole of carbonate and 0.5 millimole of hydrogencarbonate. Here again, problem should read: "The second end point was reached when AN EXTRA 20 ml of the acid was added

  12. Aadarsh
    • 2 years ago
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    Oooookkkkkk, I will consult with my teacher about this. But these r the exact wordings that our teacher gave.

  13. Vincent-Lyon.Fr
    • 2 years ago
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    What about the answers?

  14. Vincent-Lyon.Fr
    • 2 years ago
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    btw, do you use 22.4 litres/mole or 22.7 litres/mole for molar volume in STP?

  15. Aadarsh
    • 2 years ago
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    22.4

  16. Vincent-Lyon.Fr
    • 2 years ago
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    Do you have expected answers ?

  17. Aadarsh
    • 2 years ago
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    Not now. i need to submit it as my assignment. i couldn't do it, so i asked it here.

  18. Vincent-Lyon.Fr
    • 2 years ago
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    Do my answers coincide with yours of Mani-Jha's? Do you understand what reactions are at stake here?

  19. Aadarsh
    • 2 years ago
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    My teacher gave the answer just now as 112 ml for the first one, and for the second one: Weight of NaOH = 0.8g, weight of Na2CO3 = 1.59 g

  20. Vincent-Lyon.Fr
    • 2 years ago
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    1st one is correct. I forgot to divide something by 5, as you take only a 20 mL sample of the 100 mL solution. So there were 5 millimoles of ammonia ie a volume of 112 mL

  21. Aadarsh
    • 2 years ago
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    Thanks a lot

  22. Vincent-Lyon.Fr
    • 2 years ago
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    I have 159 mg of Na2CO3 which is not 1.59 g, unless concentration af acid is 1M instead of 0.1M. Then I do not understand the NaOH part. Your solution is supposed to have NaHCO3. What do you think?

  23. Aadarsh
    • 2 years ago
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    Even I think the same.

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